Problem: The lifespans of meerkats in a particular zoo are normally distributed. The average meerkat lives $10.7$ years; the standard deviation is $1.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a meerkat living less than $12.1$ years.
$10.7$ $9.3$ $12.1$ $7.9$ $13.5$ $6.5$ $14.9$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $10.7$ years. We know the standard deviation is $1.4$ years, so one standard deviation below the mean is $9.3$ years and one standard deviation above the mean is $12.1$ years. Two standard deviations below the mean is $7.9$ years and two standard deviations above the mean is $13.5$ years. Three standard deviations below the mean is $6.5$ years and three standard deviations above the mean is $14.9$ years. We are interested in the probability of a meerkat living less than $12.1$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the meerkats will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the meerkats will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $9.3$ years and the other half $({16\%})$ will live longer than $12.1$ years. The probability of a particular meerkat living less than $12.1$ years is ${68\%} + {16\%}$, or $84\%$.